\(\frac{1}{1.3}+\left|\frac{-1}{3.5}\right|+\left|\frac{-1}{5.7}\right|+...+\left|\frac{-1}{49.51}\right|=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)=\frac{1}{2}\left(1-\frac{1}{51}\right)=\frac{1}{2}.\frac{50}{51}=\frac{25}{51}\)

    \(\frac{1}{1.3}+\left|-\frac{1}{3.5}\right|+\left|-\frac{1}{5.7}\right|+...+\left|-\frac{1}{49.51}\right|\)

\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49\cdot51}\)

\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{51}\right)\)

\(=\frac{1}{2}.\frac{50}{51}\)

\(=\frac{25}{51}\)

trả lời chi tiết cho mk nhé mb

cs sai đề bài ko ạ

Ta có : \(S=3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^8}+\frac{3}{2^9}\)

=> 2S = \(6+3+\frac{3}{2}+...+\frac{3}{2^7}+\frac{3}{2^8}\)

=> 2S - S = \(\left(6+3+\frac{3}{2}+...+\frac{3}{2^7}+\frac{3}{2^8}\right)-\left(3+\frac{3}{2}+\frac{3}{2^2}+...+\frac{3}{2^8}+\frac{3}{2^9}\right)\)

=> S = \(6-\frac{3}{2^9}\)

\(\text{a) }\text{Vì }\hept{\begin{cases}\text{x ⁝ 35}\\\text{x ⁝ 63}\\\text{x ⁝ 105}\end{cases}\Rightarrow}\text{x ∈ BC( 35 , 63 , 105 )}\)

\(\text{Vì :}\)

35 = 5 . 7

63 = 3² . 7

105 = 3 . 5 . 7

=> BCNN ( 35 , 63 , 105 ) = 32 . 5 . 7 = 315

=> BC( 35 , 63 , 105 ) = { 315 ; 630; 945 } ( do các phần  tử cói 3 chữ số )

\(\text{Giải :}\)

\(\text{a) }\text{Vì }\hept{\begin{cases}\text{x ⁝ 35}\\\text{x ⁝ 63}\\\text{x ⁝ 105}\end{cases}\Rightarrow}\text{x ∈ BC( 35 , 63 , 105 )}\)

\(\text{Vì :}\)

35 = 5 . 7

63 = 3² . 7

105 = 3 . 5 . 7

=> BCNN ( 35 , 63 , 105 ) = 32 . 5 . 7 = 315

=> BC( 35 , 63 , 105 ) = { 315 ; 630; 945 } ( do các phần  tử cói 3 chữ số )

Ta có dãy:

\(0^2,1^2,2^2,3^2,4^2,5^2,6^2,7^2,8^2,9^2,10^2,11^2,12^2,13^2,14^2,15^2,16^2,17^2,18^2,19^2,20^2\)

\(\Rightarrow0,1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400\)

\(x\)chia hết cho \(35,63,105\)nên \(x\)là \(BC\left(35,63,105\right)\).

Phân tích thành tích các thừa số nguyên tố: 

\(35=5.7,63=3^2.7,105=3.5.7\).

\(BCNN\left(35,63,105\right)=3^2.5.7=315\)

suy ra \(x\in B\left(315\right)\)mà \(x\)là số có ba chữ số nên \(x\in\left\{315,630,945\right\}\).

111 số nha. k mình với

Đáp án:

Từ 1 đến 999 có các số chia hết cho 9 là :

(999 - 9) : 9 + 1 = 111 ( số )

Hok tốt

\(E=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2009^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{2009.2009}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2020}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2009}-\frac{1}{2010}=1-\frac{1}{2010}< 1\)

\(2xTS=2+2^2+2^3+2^4+...+2^{2007}\)

\(\Rightarrow TS=2xTS-TS=2^{2007}-1\)

\(\Rightarrow B=\frac{2^{2007}-1}{1-2^{2009}}\)